How do you evaluate the definite integral #int (sinx+cosx)^2dx# from #[pi/6,pi/4]#?

1 Answer
Noah G
Jan 3, 2017

#int_(pi/6)^(pi/4)(sinx + cosx)^2dx = (pi + 3)/12 ~~ 0.51#

Explanation:

Expand:

#=int_(pi/6)^(pi/4)(sin^2x + cos^2x + 2sinxcosx)dx#

Apply the identity #sin^2x + cos^2x = 1#:

#=int_(pi/6)^(pi/4) (1 + 2sinxcosx)dx#

Apply the identity #sin2x = 2sinxcosx#:

#=int_(pi/6)^(pi/4) (1 + sin2x)dx#

Let #u= 2x#. Then #du = 2dx# and #dx= 1/2du#.

#=int_(pi/6)^(pi/4) (1 + sinu) * 1/2du#

Use the property #int(af(x)) = aintf(x)# where #a# is a constant.

#=1/2int_(pi/6)^(pi/4) (1 + sinu)du#

Make sure you know your basic integrals of trig functions (e.g. sine and cosine). Also, it would be good to commit to memory the formula #int(x^n)dx = x^(n + 1)/(n + 1) + C#. #color(red)("Remember")#: we are integrating with respect to #u#.

#=1/2[u - cosu]_(pi/6)^(pi/4)#

Since #u = 2x#:

#=1/2[2x - cos(2x)]_(pi/6)^(pi/4)#

Evaluate using #int_a^bF(x) = f(b) - f(a)#, where #f'(x) = F(x)#.

#=1/2[2(pi/4) - cos(2(pi/4))] - 1/2[2(pi/6) - cos(2(pi/6))]#

#=1/2[pi/2 - 0] - 1/2[pi/3 - 1/2]#

#=pi/4 - pi/6 + 1/4#

#= (pi + 3)/12#

#~~0.51#

Hopefully this helps!

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